How to find motion equations using energy and angular momentum? [closed]

Question

I'd really appreciate some help with an exercise. The exercise presents a system of two bodies, $m$ and $M$. Both are connected with a weightless rope; the former is rotating (with a given initial tangential velocity of $v_0$) on a friction-less table, at an initial distance of $r_0$. The latter is hanging through a hole in the table (the setting is depicted in the attached picture). Considering all, how can I get the motion equations ($r(t)$) of the system using the conservation of energy and angular momentum?

Here's the answer I've reached using the conservation of angular momentum at the point of rotation (the hole), and then the conservation of energy. When asked to find the motion equation, is it enough to do as shown in the picture? Or must I find a concrete relation between r and t, and if so, what other equations can I use?:

Answer 1

Just consider forces on both bodies. On the first using acceleration in polar coordinates we get $$-T\hat{r} = m\vec{a} = m\left((\ddot{r} -r\omega^2)\hat{r} + (2\dot{r}\omega + r\dot{\omega})\hat{\phi}\right)$$ so $2\dot{r}\omega + r\dot{\omega} = 0$ which when multiplied by $r$ gives $\frac{d}{dt}(r^2\omega) = 0$, which is precisely your conservation of angular momentum $r^2\omega = r_0^2\omega_0$.

The $\hat{r}$ direction gives $-T = m(\ddot{r}-r\omega^2)$. The length of the rope is constant so the acceleration of the second body is $-\ddot{r}$ which implies $-M\ddot{r} = -T+Mg$. Now eliminate $T$ and $\omega$ using $r^2\omega = r_0^2\omega_0$ and you have your equation of motion.

Answer 2

From angular momentum, you can get, v, as a function of, r. If you include the K.E. Of the dropping mass, M, in the energy equation, you can get an expression for, dr/dt. Integrate that to get r(t).

Answer 3

What do you necessarily want to find? If you want to find all the equations, then you do not need to use angular momentum.

Initially, calculate E.O.M for the lower 'M', by using the equation of motion

mg - T = ma

If it is given that acceleration is 0, then just substitute a = 0 there.

Then, for the upper body, it is rotating with a speed of v, so mv^2 / r would be the outward force acting on it.

Equate mv^2/r = T, for the small mass 'm' to remain in a circle.

Hope it helped!

Answer 4

Initially, before M starts to drop, m has only a tangential velocity $v_0$. Once M starts to fall, m has both a tangential and a radial velocity that must be considered, easiest to evaluate using polar coordinates. You can obtain the differential equation (de) of motion using the conservation of angular momentum and energy balance. If you cannot manually solve the de, it can be solved using Mathematica or another high end math software program.

You can also solve for the tension in the rope using $F = ma$ on both m and M.

This problem can also be solved using Lagrange's equations.